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jek_recluse [69]
3 years ago
6

HOW DOES DENSITY AFFECT THE MELTING POINT OF WATER?

Chemistry
1 answer:
ddd [48]3 years ago
5 0
<span> As a liquid is heated, its vapor pressure increases until the vapor pressure equals the pressure of the gas above it. Bubbles of vaporized liquid (i.e., gas) form within the bulk liquid and then rise to the surface where they burst and release the gas. (At the boiling temperature the vapor inside a bubble has enough pressure to keep the bubble from collapsing.) In order to form vapor, the molecules of the liquid must overcome the forces of attraction between them.<span> The temperature of a boiling liquid remains constant, even when more heat is add.</span></span>
You might be interested in
Fill in the blanks (Balancing chemical reactions)<br>BRAINLIEST
Artemon [7]

\huge{ \mathfrak{  \underline{ Answer \:  \:  ✓ }}}

The 2 in front of Na in 2Na + Cl_2 \rightarrow  2NaCl is <u>Coefficient.</u>

_____________________________

The _2 after Cl_2 in the same equation is <u>sub </u><u>script</u><u>.</u>

_____________________________

There are two atoms of Na in 2NaCl

_____________________________

\mathrm{ \#TeeNForeveR}

7 0
3 years ago
At 338 mm hg and 72 c a sample of carbon monoxide gas occupies a volune of 0.225 L the gas transferred to a 1.50 L flask and the
Elis [28]

Answer:

P₂  = 0.09 atm

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 0.225 L

Initial pressure = 338 mmHg (338/760 =0.445 atm)

Initial temperature = 72 °C (72 +273 = 345 K)

Final temperature = -15°C (-15+273 = 258 K)

Final volume = 1.50 L

Final pressure = ?

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₂ = P₁V₁ T₂/ T₁ V₂ 

P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L

P₂  =  25.83 atm .L.  K  / 293 K . L

P₂  = 0.09 atm

7 0
3 years ago
Please help with 3! Please give only the correct answer...
cupoosta [38]
The answer is:  " 1.75 * 10 ^(-10)  m " .
_________________________________________________________
Explanation: 
_________________________________________________________
This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
_________________________________________________________
Given: " 0.000000000175 m " ;  write this in "scientific notation.
_________________________________________________________
Note:   After the "first zero and the decimal point" {Note: that first zero that PRECEDES the decimal point in merely a "placeholder" and does not count as a "digit" — for our purposes} —
                     There are NINE (9) zeros, followed by "175"
_______________________________________________________
To write in "scientific notation", we find the integer that is written, as well, as any "trailing zeros" (if there are any—and by "trailing zeros", this means any number consecutive zeros/and starting with "the consecutive zeros" only —whether forward (i.e., "zeros following"; or backward (i.e. "zeros preceding").

In our case we have "zeros preceding";  that is a decimal point with zeros PRECEDING an "integer expression"<span>
</span><span> (the "integer" is "175").</span>
______________________________________________________
We then take the "integer expression" (whatever it may be:  12, 5, 30000001 ; or could be a negative value,  etc.) ;  

→  In our case, the "integer expression" is:  "175" ;

and take the first digit (if the expression is negative, we take the negative value of that digit;  if there is only ONE digit (positive or negative), then that is the digit we take ;

And write a decimal point after that first digit (unless in some cases, there is only one digit);  and follow with the rest of the consecutive digits of that 'integer expression' ;

→ In our case:  "175" ; becomes:  " 1.75" .
__________________________________________________
Then we write:  "  * 10^ "
__________________________________________________
   {that is "[times]"; or "multiplied by" :    [10 raised exponentially to the power of  <u>     </u> ]._____________________________________________________
 And to find that power, we take the "rewritten integer value (i.e. "whole number value that as been rewritten to a single digit with a decimal point"); and count the [number of "trailing zeros";  if there are any; PLUS the number of decimal places one goes] ; and that number is the value to which "10" is raised.
{If there are none, we write:  " * 10⁰ " ;    since "any value, raised to the "zero power", equals "1" ; so " * 10⁰ " ; is like writing:  " * 1 " .

If there are "trailing zeros" AND/OR or  any number of decimal places,  to the "right" of this expression; the combined number of spaces to the right is: 
  { the numeric value (i.e. positive number) of the power to which "10" is raised }.

Likewise, if there are "trailing zeros" AND/OR or any number of decimal places, to the "LEFT" of this expression; the combined number of spaces to the LEFT is the value of the power which "10" is raised to; is that number—which is a negative value.

In our case:  we have:  0.000000000175 * 10^(-10) .

Note:  The original notation was:

             →  " 0.000000000175 m "

{that is:  "175" [with 9 (nine) zeros to the left].}.

We rewrite the "175" ("integer expression") as:

"1.75" .
____________________________________________________
So we have:
         →   " 0.000000000175 m " ;

Think of this value as:

        " 0. 0000000001{pseudo-decimal point}75   m ".

And count the number of decimal spaces "backward" from the
      "pseudo-decimal point" to the actual decimal; and you will see that there are "10" spaces (to the left).   
______________________________________________________
Also note:  We started with "9 (nine)" preceding "zeros" before the "1" ;  now we are considering the "1" as an "additional digit" ;
             →  "9 + 1 = 10" .
______________________________________________________
Since the decimals (and zeros) come BEFORE (precede) the "175" ; that is, to the "left" of the "175" ; the exponent to which the "10" is raised is:
 "NEGATIVE TEN" { "-10" } .

So we write this value as:  " 1.75 * 10^(-10)  m " .  

{NOTE:  Do not forget the units of measurement; which are "meters" —which can be abbreviateds as:  "m" .} . 
______________________________________________________
The answer is:  " 1.75 * 10^(-10)   m " .
______________________________________________________
4 0
3 years ago
The molecular structure of water contains two atoms of hydrogen and one atom of oxygen. When water reaches its boiling point and
Andrews [41]
It turns from a liquid to a gas
7 0
3 years ago
Which of these substances has the lowest pH? 0.5 M HBr, pOH = 13.5 0.05 M HCl, pOH = 12.7 0.005 M KOH, pOH = 2.3
vagabundo [1.1K]
<h3><u>Answer;</u></h3>

0.5 M HBr, pOH = 13.5 ; Has the lowest pH

<h3><u>Explanation;</u></h3>

From the question;

pH = -Log [OH]

or pH = 14 - pOH

Therefore;

For 0.5 M HBr

[H+] = 0.5 M

pH = - Log [0.5]

     = 0.30

For;  pOH = 13.5

pH = 14 - pOH

     = 14 -13.5

     = 0.5

For; 0.05 M HCl

pH = - log [H+]

[H+] = 0.05

pH = - Log [0.05]

     = 1.30

For; pOH = 12.7

pH = 14 -pOH

     = 14 -12.7

     = 1.30

For;  0.005 M KOH,

pOH = - log [OH]

[OH-] = 0.005

pOH = - Log 0.005

        = 2.30

pH = 14 - 2.30

     = 11.7

For; pOH = 2.3

   pH = 14 -pOH

         = 14- 2.3

         = 11.7

6 0
3 years ago
Read 2 more answers
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